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If p+q+r=a+b+c=0 , then the determinant Δ=pagbrcqcrapbrbpcqa equal 

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a
0
b
1
c
pa+qb+rc
d
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detailed solution

Correct option is A

we have Δ=pqra3+b3+c3-abcp3+q3+r3But a+b+c=0⇒(a+b)3=-c3⇒  a3+b3+3ab(a+b)+c3=0⇒  a3+b3+c3=-3ab(-c)=3abcSimilarly, p3+q3+r3=3pqr Thus , Δ=pqr(3abc)-abc(3pqr)=0


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